Question

$ABCD$ is a quadrilateral in which $AB$ and $CD$ are smallest and longest sides respectively.
Prove that $\angle A>\angle Cand\angle B>\angle D.$

Answer 1

REF.image

Given

AB is the smallest side -----(1)

CD is the largest side -----(2)

To prove $:<A>\angle C,<B>\angle D$

Proof:

Join A and c,and mark the angles

In $△ABC,$

$\therefore BC>AB$$.................\left(from\right(1\left)\right)$

$\therefore \angle 1=\angle 2$

Angle opposite to the longer side greater.

In$△ADC,$

$CD>AD.....................\left(from\right(2\left)\right)$

$\therefore \angle 3=\angle 4$

Angle opposite to the longer side is greater.

Adding equation

$\angle 1+\angle 3>\angle 2+\angle 4$

Now ,solving for $\angle B>\angle D$

Join B and d, and mark the angles.

In $△ABD$,

$AD>AB...................\left(from\right(1\left)\right)$

$\therefore \angle 5=\angle 6$

In $△CBD,$

$CD>BC...............\left(from\right(2\left)\right)$

$\therefore \angle 7=\angle 8$

Adding equations

$\angle 5+\angle 7>\angle 6+\angle 8$

$\Rightarrow \angle B>\angle D$

Hence proved.